Now we isolate the variable Q. Since Q is multiplied by –0.2, we divide both sides of the equation by–0.2.
After completing this unit you should be able to:
In the previous unit, we have had the convenience of only having one variable in each equation. In the real world we often have many variables in a single equation. These equations containing more than one variable are referred to as "multivariate" equations.
When faced with an equation with more than one variable, you may either wish to find a numeric value for each variable, or solve for one variable in terms of the other. In this unit you will review how to solve for one variable in terms of others. This means you are isolating one variable and your result will contain other variables. How you approach solving multivariate equations will depend on whether you intend to find a numeric value for each variable, or just solve for one variable.
When we are given a multivariate equation where we want to solve for just one variable, we follow the same steps used for equations with one variable as outlined in the last unit.
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Let’s look at an example presented earlier in the book. In this example, the equation of the price (P) of a paperback book is given in terms of the quantity sold (Q). The following equation represents P in terms of Q.
What if we wish to know the quantity sold in terms of the price? In this case, we would have to solve the equation for Q, the quantity of books sold, meaning we want to isolate Q. We follow the steps above to do this.
1. Combine like terms.
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If we look at the equation we can see that like terms are already combined so we can skip to step 2. |
2. Isolate the terms that contain the variable you wish to solve for.
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Use the Properties of Addition, and/or Multiplication and their inverse operations to isolate the terms containing the variable you wish to solve for. |
P= -0.2(Q-1)+7 P= -0.2Q+0.2+7 P= -0.2Q+7.2 P-7.2= -0.2Q+7.2 -7.2 P-7.2= -0.2Q |
3. Isolate the variable you wish to solve for.
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Now we isolate the variable Q. Since Q is multiplied by –0.2, we divide both sides of the equation by–0.2. |
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4. Substitute your answer into the original equation and check that it works.
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Checking an answer with this type of solution can be a bit more complicated, but we follow the same procedure we do with numeric solutions. However, we do not need to have a numerical value to use the substitution technique. We can also use an expression that contains variables. In the example: |
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Further Thought: Look closely at the example above. Can you tell why the above equation could be a problem for our book seller? |
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Let’s look at another example of solving a multivariate equation.
Given the equation 2(5x + z) = 30x + 3y + 10, find the value of x in terms of y and z.
Given the equation 2(5x + z) = 30x
+ 3y + 10, find the value of x in terms of y
and z.
The value of x is:
Given the equation 2(5x + z) = 30x + 3y + 10, find the value of x in terms of y and z. To solve this we should work through the steps below.
1. Combine like terms.
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First we need to expand the equation by multiplying out the parentheses. Then we combine the x terms in the equation. To do this, it's easiest to subtract 10x from both sides since this leaves us with a positive value of x |
2(5x + z)=30x + 3y + 10 10x + 2z = 30x + 3y + 10 10x - 10x + 2z = 30x - 10x + 3y + 10 2z = 20x + 3y + 10 |
2. Isolate the terms that contain the variable you wish to solve for.
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Now we want to isolate the term with the x variable. To do this we subtract 3y and 10 from both sides of the equation and combine terms. |
2z - 3y - 10 = 20x + 3y - 3y + 10 - 10 2z - 3y - 10 = 20x |
3. Isolate the variable you wish to solve for.
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Since the term that contains x is multiplied by 20, we divide both sides of the equation by 20 to isolate x. |
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4. Substitute your answer into the original equation and check that it works.
  When substituting the value for Q back into our original equation we found that both sides of the equation are equal, so this solution is correct. |
Now take time to practice solving multivariate equations that contain a single variable by completing the practice before moving on to the next unit.
Further Thought Response: For the equation P = (–0.2)(Q – 1) + 7, for values of Q = 36 the price of the book is zero. For values of Q greater than 36 the bookseller has to pay people to take the books (P has a negative value).
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